2.6 Forced Oscillations And Resonance

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2.6Forced Oscillations and Resonance1Oscillator equation with external force F (t): basic case assumes F periodic,mx 00 cx 0 kx F0 cos ωtMany real-life situations can be modelled with this equation, for example buildings in an earthquake.There are three standard cases.Case 1: BeatingqTake c 0 (no damping/friction) and ω 6 ω0 mk (driving frequency 6 natural frequency).Already found complementary function xC (t) c1 cos ω0 t c2 sin ω0 t.Particular integral: guess x P (t) a cos ωt b sin ωt. Thenmx 00P kx P ( maω 2 ka) cos ωt ( mbω 2 kb) sin ωtF0F0 F0 cos ωt a , b 022k mωm ( ω0 ω 2 )F0 x P (t) cos ωt2m ( ω0 ω 2 )F0cos ωt c1 cos ω0 t c2 sin ω0 t x (t) x P (t) xC (t) 2m ( ω0 ω 2 ) Sum of distinct periodic motions. Larger F0 more motion. ω close to ω0 more motion.Suppose have initial conditions x (0) 0 x 0 (0) (periodic force applied to resting spring). Quicklyobtain 2F0ω0 ωF0ω0 ωcosωt cosωt sinx (t) t sint022m(ω02 ω 2 )m(ω02 ω 2 )using a trigonometric identity. If ω0 , ω close in value, then ω0 ω ω0 ωAmplitude beats at ω02 ω rad/s.x0xω0 2024ω 1868Graphics show x 00 400x 38 cos 18t where ω0 20. Solution x (t) sin t sin 19t A(t) sin 19t.High frequency vibration sin 19t with periodic amplitude A(t) sin t.1 Thisis an abstract summary. Study this open-book and pay attention to the numerical examples from lectures. . .10t

Case 2: ResonanceNo friction (c 0) where ω ω0 Driving frequency natural frequencyObvious guess x P (t) a cos ω0 t b sin ω0 t already solves homogeneous ODE, so tryx P (t) at cos ω0 t bt sin ω0 tSubstituting in the ODE and solving for a, b givesx P (t) F0t sin ω0 t2mω0x P is also the solution corresponding to2 the initial conditions x (0) x 0 (0) 0. Sine wave with increasing amplitude A(t) F0t2mω0 Spring tears itself apart!Example Consider the IVP(4x 00 64x 4 cos ωtx (0) x 0 (0) 0 The natural frequency is ω0 k/m 4 rad/s. The four graphs show the solutions for fourdifferent driving frequencies: the last case is resonance.xx0.0050.102ω 20x0.10x (t) 68010t1sin 8t sin 12t192x2ω 6468x (t) 1sin t sin 5t1010t422ω 52 It44x (t) 68010t 22t9tsin sin922 4is also what we obtain by taking limω ω0 x (t) from the previous slide22ω 446x (t) 81t sin 4t810t

Summary of undamped-driven motion The initial value problem mx 00 kx F0 cos ωt with initialconditions x (0) 0 x 0 (0) has solution 2F0 /mω0 ωω0 ω sint sint if ω 6 ω0 2222ω0 ωx (t) F 0 t sin ω0 tif ω ω02mω0 where ω0 k/m. As ω ω0 low frequency beats of increasing amplitude occur.The clickable animation is generated with the same F0 , m throughoutxω0 50ω 48246810tCase 3: Damped-driven motion (practical resonance)c 0x 00 2px 0 ω02 x mx 00 cx 0 kx F0 cos ωtF0cos ωtmTransient and steady-periodic solutions Three types:DampingOverdampingCondition Complementary Function xC (t) 2 2 2 2 c2 4km e pt c1 e p ω0 t c2 e p ω0 tCritical damping c2 4kmUnderdampingc2 4km(c1 c2 t)e pte pt c1 cos ω1 t c2 sin ω1 t where ω1 p 0 xC transient: lim xC (t) 0qω02 p2t Particular integral: standard guess x P (t) a cos ωt b sin ωt always works. Regardless of initial conditions, x (t) x P (t) for large t: steady-periodic solution.Example Find the steady-periodic solution to x 00 3x 0 2x cos ωt.Try x P (t) a cos ωt b sin ωt. Substitute in the ODE:( aω 2 3bω 2a) cos ωt ( bω 2 3aω 2b) sin ωt cos ωtω2 23ω, b 22229ω (ω 2)(ω 2)2 9ω 2 11cos (ωt γ) x P (t) (ω 2 2) cos ωt 3ω sin ωt 2222(ω 2) 9ω(ω 2)2 9ω 2 a where γ is the phase angle.3

General situation Long-term solution is F0 /m(ω02 ω 2 ) cos ωt 2pω sin ωt222 ω ) (2pω )F0cos(ωt γ) qm (ω02 ω 2 )2 (2pω )2x P (t) (ω02where tan γ 2pω ω2ω02 Amplitude is a function of frequency ω. Maximum amplitude when denominator minimal: a little calculus shows this is when(q(requires very light damping c2 2km)ω02 2p2 if 2p2 ω02ω 0if 2p2 ω02In the first situation this is known as practical resonance.The animation shows the steady periodic solution for the equation x 00 16x 0 324x cos ωt, namelyx P (t) p1(182 ω 2 )2 (16ω )2cos(ωt γ)for different values of driving frequency ω. Note that the practical resonant frequency (maximumamplitude) occurs when ω 14.xω0 182qω02 2p2 144ω 164810t

Glass smashing!Model transverse motion of lip of a wine glass by31 0010 x 15 x 0 1, 000, 000x F0 cos ωtF0 cos ωt models vibration of the air due to ambient sound.Unforced motion (F0 0) of the glass isxxC (t) e t (c1 cos ω1 t c2 sin ω1 t)whereωω1 12π2πqω1ω02 p2 2πq0km c24m2 503.2920959 Hz12( 1 octave above middle C)Practical resonance occurs whenqω1f ω02 2p2 503.2920707 Hz2π2πThe animation shows the steady periodic solution for tiny variations of driven frequency f near thepractical resonant frequency.xf Practical Resonant Frequency 5 Hz0.1tA singer must be very loud accurate to crack the glass.3 Smallmass, small damping, high spring constant53t

Application: Electrical circuitsAn RLC circuit has a source voltage4 V (t) V0 sin ωt, a resistor R ohms, a capacitor C farads, andan inductor L henries connected in series.Current flow: I dQdt where I ( t ) current flow (amps) and Q ( t ) (coulombs) is the charge stored inthe capacitor at time tCalculate voltage drop across each component: RI,L1C Q,L dIdt respectivelyODEdI1d2 IdI1 RI Q V (t) L 2 R I V 0 (t)dtCdtdtCDamped-driven spring equation in disguise, withm Lk C 1c RF0 ωV0Electrical Resonance Amplitude of steady-periodic current I qMaximal when ω 2 LC 1applications in electronics. . .V0R2 ωL 1 2ωCAudio Hum Background ‘noise’ often result of (practical) resonant current resonanceSolution: adjust capacitance C to be very different to ω 2 L 1 to minimize resonant current.Reducing power loss If C 0, the current flow isI (t) V0R2 ω 2 L2cos(ωt γ)Inductor reduces current flow: acts like extra resistance. Inserting capacitor C ω 2 L 1 increases current to V0 /R: reduces power loss from natural inductance.Tuning an (AM) radio Radio station frequency ω induces voltage V (t) V0 sin ωt in a radio antenna, resulting in a current flow which (after amplification) powers a loudspeaker.Variable capacitor tuned to C ω 2 L 1 amplifies signal at frequency ω and diminishes currentflow due to other radio frequencies.Analogue radio tuning dials were variable capacitors: changing C changed the resonant frequency, and thus which radio station was amplified most.4 In US, mains electricity is V (t) 110 2 sin(120πt)6

2.6 Forced Oscillations and Resonance1 Oscillator equation with external force F(t): basic case assumes F periodic, mx00 cx0 kx F0 coswt Many real-life situations can be modelled with this equation,

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